Try this beautiful problem from Geometry based on the Area of the Trapezium
Area of the Trapezium – AMC-10A, 2018- Problem 24
Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$. Let $D$ be the midpoint of $\overline{AB}$, and let $E$ be the midpoint of $\overline{AC}$. The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$, respectively. What is the area of quadrilateral $FDBG$?
- $79$
- $ 75$
- $82$
Key Concepts
Geometry
Triangle
Trapezium
Check the Answer
But try the problem first…
Answer: $75$
AMC-10A (2018) Problem 24
Pre College Mathematics
Try with Hints
First hint

We have to find out the area of BGFD.Given that AG is the angle bisector of \(\angle BAC\) ,\(D\) and \(E\) are the mid points of \(AB\) and \(AC\). so we may say that \(DE ||BC\) by mid point theorm…
So clearly BGFD is a Trapezium.now area of the trapezium=\(\frac{1}{2} (BG+DF) \times height betwween DF and BG\)
can you find out the value of \(BG,DF \) and height between them….?
Can you now finish the problem ……….
Second Hint

Let $BC = a$, $BG = x$, $GC = y$, and the length of the perpendicular to $BC$ through $A$ be $h$.
Therefore area of \(\triangle ABC\)=\(\frac{ah}{2}\)=\(120\)………………..(1)
From the angle bisector theorem, we have that\(\frac{50}{x} = \frac{10}{y}\) i.e \(\frac{x}{y}=5\)
Let \(BC\)=\(a\) then \(BG\)=\(\frac{5a}{6}\) and \(DF\)=\(\frac{1}{2 } \times BG\) i.e \(\frac{5a}{12}\)
now can you find out the area of Trapezium ?
can you finish the problem……..
Final Step

Therefore area of the Trapezium=\(\frac{1}{2} (BG+DF) \times FG\)=\(\frac{1}{2} (\frac{5a}{6}+\frac{5a}{12}) \times \frac{h}{2}\)=\(\frac{ah}{2} \times \frac{15}{24}\)=\(120 \times \frac{15}{24}\)=\(75\) \((from ……..(1))\)
Other useful links
- https://www.cheenta.com/angles-in-a-circle-prmo-2018-problem-80/
- https://www.youtube.com/watch?v=V01neV8qmh4